Billiard ball collision with change in direction

[mariac]

Homework Statement

A billiard ball with mass 0.210 kg has a speed of 5.2 m/s and collides completely inelastically with another ball of the same size and mass.

A. Find the final speed of the two balls.
The answer I got was 2.6 m/s. The next part is trickier.

B. Another billiard ball traveling at 6.25 m/s collides with a second stationary ball of equal size and shape causing the second ball to move at an angle of 25 degrees with respect to the horizontal at a speed of 3.0 m/s.

What impulse (magnitude and direction) did the each ball receive? What is the final velocity (magnitude and direction ) of the first ball?

Homework Equations

The Attempt at a Solution

Conceptually, I don't understand why the second ball would move at an angle. Would that mean that the first ball hits the second ball off-center? I know that the total momentum must remain the same, and p=6.25m/s x 0.210kg = 1.31kg m/s. Would this be the magnitude of the impulse received by the second ball? Or would the impulse be only half of this since the first ball receives an equal impulse? If the first ball hit the second slightly off center, this would change the direction of the impulse. I presume the first ball would turn 25 degrees below horizontal after the collision. Would the first ball proceed at a velocity of 3.25 m/s, or would the change in direction somehow affect the velocity?
I'm really confused. Can you help?

 

[krausr79]

I don't have much experience w/ impulse, but your guesses seem correct. It would be an off-center hit to cause this. To find out the direction/magnitude of the first ball, break it into x-y components. In both an x and y sense seperately, Start momenta added together should equal final momenta added together. You can find out the starting momenta of both balls, and the final momentum of the second ball; which will allow you to solve for the final momentum of the second ball. Done in x and y you will have components, then you can switch it back to polar (direction, magnitude).

 

[thomate1]

I would approach this problem by using the conservation of momentum and the conservation of energy principles. In a perfectly elastic collision, the total momentum and kinetic energy of the system would remain constant. However, in this case, the collision is completely inelastic, which means that some of the kinetic energy is lost and converted into other forms of energy, such as heat and sound.

To answer part A, we can use the conservation of momentum principle, which states that the total momentum before the collision is equal to the total momentum after the collision. In this case, the initial momentum is 0.210 kg x 5.2 m/s = 1.092 kg⋅m/s. Since the balls stick together after the collision, their final mass is 0.210 kg + 0.210 kg = 0.420 kg. Therefore, the final velocity of the combined balls can be calculated as 1.092 kg⋅m/s ÷ 0.420 kg = 2.6 m/s.

For part B, we can use the same principle and apply it to each ball separately. The initial momentum of the first ball is 0.210 kg x 6.25 m/s = 1.3125 kg⋅m/s. After the collision, the first ball will have a final velocity of 3.0 m/s at an angle of 25 degrees with respect to the horizontal. This means that the horizontal component of its velocity will be 3.0 m/s x cos(25) = 2.72 m/s, and the vertical component will be 3.0 m/s x sin(25) = 1.28 m/s. Therefore, the change in velocity for the first ball is (2.72 m/s, 1.28 m/s).

For the second ball, we can use the same approach and find that its final velocity will be (3.0 m/s, -1.28 m/s) since it is moving at an angle of 25 degrees below the horizontal. The change in velocity for the second ball is (3.0 m/s, -1.28 m/s).

To calculate the magnitude and direction of the impulse received by each ball, we can use the impulse-momentum theorem, which states that impulse is equal to the change in momentum. Therefore, the impulse received by the first ball is (1.3125